>HI Gregory . Thanks for replying.
>OK if both results are integers but will it do any rounding
>e.g. 201106/ 100 = 2011.06 Integer part = 2011 Fine


If both operands are integer, there is no rounding : http://msdn.microsoft.com/en-us/library/3b1ff23f.aspx
When you divide two integers, the result is always an integer. For example, the result of 5 / 2 is 2.

>If I have both integers say 201199 / 100 = 2011.99, what result will C# give ? 2011 or 2012 ?

201199 / 100 yields 2011


>(It seems that the Default ROUND is not the 'Normal' round expected (i.e. 6.5 gos to 6 instead of 7) and wondering if just integers used
>without any rounding parameter is there any implication though with rounding ?)


I do not understand that question


>
>Regards,
>Gerard
>
>
>>>Is there an integer Function in C#
>>>
>>>(VFP Code)
>>>Int(201106 / 100) ....... Returns 2011
>>>Int(201107 / 100) ........ Returns 2011
>>>
>>>(In above examples , 201106 and 201107 are both integers
>>>
>>>Tia
>>>Gerard
>>
>>
>>If both operands ( 201106 and 100) are integer then the result will be integer -
>>
>>Otherwise you can
>> - use Math.Floor(201106.0 / 100.0)
>> - cast it to integer
>>

>>  int result = (int)(201106.0 / 100.0)
>>